Register to reply 
What would E field addition be if Gauss's law was different 
Share this thread: 
#1
Feb1713, 05:05 PM

P: 757

What if Gauss' Law was
div(E)=q^{2}/eps_{0} Then if we had two point charges, how could we calculate the resultant E field at an arbitrarily point? Obviously superposition would not work anymore, so how could it be done mathematically? This is essentially a math question 


#2
Feb1713, 05:15 PM

C. Spirit
Sci Advisor
Thanks
P: 5,620

Did you stop to check if it was first dimensionally correct before proceeding? The units for the real gauss's law pertain to a physically interpretable / measurable quantity.



#3
Feb1713, 10:18 PM

P: 757

math doesn't care about units



#4
Feb1713, 10:41 PM

P: 1,066

What would E field addition be if Gauss's law was different
I'm trying to understand your question but I don't quite follow? The right side seems to have units of charge squared divided by farads per meter, but the lefts units are completely different. What is this supposed to mean?



#5
Feb1813, 05:01 AM

Mentor
P: 16,356

You have received your answer: what you wrote does not have the correct units, so there is no answer to your question. It would be like asking "how long would it take a ball to fall 7 gallons" 


#6
Feb1913, 12:14 AM

P: 757

I guess another way to ask this is to say if a particle of special type of matter had the property that it created a vector field given by
F=(c*m^2/r°r)*r/r where m is the mass of the particle, then if I have two particles, each of mass m, one at the origin and one at x=1, then what would be the resultant field F at an arbitrary point r_{0} ??? 


#7
Feb1913, 02:26 AM

P: 25

I think you already answered your question regarding the force, since you explicitly stated it.
However, if I'm understanding you correctly, I think you might be getting at the concept of linearity and the superposition of fields. Is your question what the resulting field would be if it varied as the square of the sum of all the masses/charges in some way? If so, you likely could not use Gauss's law at all, since it's a sum which assumes that the total field can be produced by summing each up piece of field, which in turn assumes a sum over each charge/mass. If the field produced by each infinitesimal source cannot be added linearly to produce the net overall field (that is, it is not independent of the other charges/masses present), then this sum does not give you the total flux, and therefore cannot give you the total mass/charge. In other words, by summing the field, you are no longer summing the total charge/mass, since they are not linearly related. To answer your original question, if the divergence of the electric field were proportional to the square of charge, then nothing would really change other than we'd probably treat the fundamental charge as its squared counterpart (and some units would need to be added or modified). It would still be linear in superposition, though. (Note: This is intuitive reasoning. Correctness not guaranteed.) 


#8
Mar813, 11:40 PM

P: 757




#9
Mar913, 12:15 AM

P: 25




Register to reply 
Related Discussions  
Gauss's Law with nonuniform Efield  Introductory Physics Homework  46  
Gauss's Law with NONuniform E field  Introductory Physics Homework  5  
Proof of closure under addition and multiplication in a field  Calculus & Beyond Homework  8  
Vector addition in electric field  Introductory Physics Homework  14  
Gauss's Law / Electric Field  Introductory Physics Homework  1 